Let P and Q be distinct points on the parabola $y^2=2x$ such that a circle with PQ as diameter passes through the vertex O of the parabola. If P lies in the first quadrant and the area of the triangle $\Delta OPQ\text{\hspace{0.17em}is\hspace{0.17em}}3\sqrt{2}$ , then which of the following is(are) the coordinates of P?

Let point P be in first quadrant and lying on parabola  $y^2=2x(\frac{a^2}{2},a).$ Let Q  be the point  $(\frac{b^2}{2},b).$ Clearly a > 0.
 
 $$\begin{gathered} \because PQ\text{\hspace{0.17em}is the diameter of circle through P, O and Q} \\ \therefore \angle POQ={90}^0\Rightarrow \frac{a}{a^2/2}\times \frac{b}{b^2/2}=-1\Rightarrow ab=-4 \\ \Rightarrow b\text{\hspace{0.17em}is negative\hspace{0.17em}}(\because a>0) \end{gathered}$$
 Given area  $\left(\Delta POQ\right)=3\sqrt{2}$
   $\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ \frac{a^2}{2}& a& 1\\ \frac{b^2}{2}& b& 1\end{array}\right|=3\sqrt{2}\Rightarrow \frac{1}{4}ab(a-b)=\pm 3\sqrt{2}\Rightarrow a-b=\pm 3\sqrt{2}$
As a is positive and b is negative, we have  $a-b=3\sqrt{2}$
 $$\begin{gathered} \Rightarrow a+\frac{4}{a}=3\sqrt{2}(\because ab=-4) \\ \Rightarrow a^2-3\sqrt{2}a+4=0\Rightarrow (a-2\sqrt{2})(a-\sqrt{2})=0 \\ \therefore a=2\sqrt{2},\sqrt{2} \end{gathered}$$
   Point P can be  $(\frac{{\left(2\sqrt{2}\right)}^2}{2},2\sqrt{2})\text{\hspace{0.17em}or}(\frac{{\left(\sqrt{2}\right)}^2}{2},\sqrt{2})$
i.e.,  $(4,2\sqrt{2})\text{\hspace{0.17em}or\hspace{0.17em}}(1,\sqrt{2})$